Consider the 2.3 MW, 690 V, 50 Hz, 1512 rpm SCIG in Case Study 3-1. The shaft of the generator iscoupled to the wind turbine through a gearbox and its stator is directly connected to the grid of 690 V/50Hz. For a given wind speed, the generator operates at the rated speed of 1516 rpm, the generatorparameters are given asTable B-1. 2.3 MW, 690 V, 50 Hz squirrel cage induction generator (SCIG)parametersGenerator TypeSCIG, 2.3 MW, 690 V, 50 HzRated Oulput Power2.30 MW1.0 pu1.0 puRated Mechanical Power2.3339 MWRated Apparent PowerRated Line-to-line VoltageRated Phase VoltageRated Stator CurrentRated Stator FrequencyRated Power Factor2.59 MVA690 V (ms)398.4 V (rms)2168 A (ms)1.0 pu1.0 pu50 Hz1.0 puRated Rotor SpeedRated Slip0.8881512 гpm-0.0081.0 puNumber of Pole Pairs214.74 kN m1.0 pu(соntinued)Rated Mechanical Torque Table B-1. ContinuedGenerator TypeSCIG, 2.3 MW, 690 V, 50 HzRated Stator Flux LinkageRated Rotor Flux LinkageStator Wir.ding Resistance, R,Rotor Winding Resistance, R,Stator Leakage Inductance, LRotor Leakage Inductance, L,,Magnetizing Inductance, LMoment of Inertia, JInertia Time Constant, HBase Flux Linkage, ApBase Impedance, Z3Base Inductance, La1.2743 Wb (rms)1.2096 Wb (rms)1.102 mn1.497 m1.0053 pu0.9539 pu0.006 pu0.008 pu0.111 pu0.111 pu3.6481 pu0.06492 mH0.06492 mH2.13461 mH1200 kg m?5.8078 sec1.2581 Wb (rms)0.1338 N1.0 pu1.0 pu1.0 pu1.0 pu0.53513 mHBase Capacitance, CB17316.17 µFNote: H - (am2/(2S3)Neglecting rotational losses of the generator, determine the followinga) The slip, and rotor mechanical and electrical speedsb) The stator and rotor currentsc) The mechanical power and torqued) The stator and rotor winding lossese) The generator efficiency and power factorf) The stator and rotor flux linkages

Consider the 2.3 MW, 690 V, 50 Hz, 1512 rpm SCIG in Case Study 3-1. The shaft of the generator is coupled to the wind turbine through a gearbox and its stator is directly connected to the grid of 690 V/50 Hz. For a given wind speed, the generator operates at the rated speed of 1516 rpm, the generator parameters are given as Table B-1. 2.3 MW, 690 V, 50 Hz squirrel cage induction generator (SCIG) parameters Generator Type SCIG, 2.3 MW, 690 V, 50 Hz Rated Output Power 2.30 MW 1.0 pu 1.0 pu Rated Mechanical Power 2.3339 MW Rated Apparent Power Rated Line to-line Voltage Rated Phase Voltage Rated Stator Current 2.59 MVA 690 V (ms) 398.4 V (rms) 2168 A (ms) 1.0 pu 1.0 pu Rated Stator Frequency Rated Power Factor 50 Hz 1.0 pu 0.888 Rated Rotor Speed Rated Slip 1.0 pu 1512 rpm -0.008 Number of Pole Pairs 2 14.74 kN m 1.0 pu (соntinued) Rated Mechanical Tarque

Consider the 2.3 MW, 690 V, 50 Hz, 1512 rpm SCIG in Case Study 3-1. The shaft of the generator is coupled to the wind turbine through a gearbox and its stator is directly connected to the grid of 690 V/50 Hz. For a given wind speed, the generator operates at the rated speed of 1516 rpm, the generator parameters are given as Table B-1. 2.3 MW, 690 V, 50 Hz squirrel cage induction generator (SCIG) parameters Generator Type SCIG, 2.3 MW, 690 V, 50 Hz Rated Output Power 2.30 MW 1.0 pu 1.0 pu Rated Mechanical Power 2.3339 MW Rated Apparent Power Rated Line to-line Voltage Rated Phase Voltage Rated Stator Current 2.59 MVA 690 V (ms) 398.4 V (rms) 2168 A (ms) 1.0 pu 1.0 pu Rated Stator Frequency Rated Power Factor 50 Hz 1.0 pu 0.888 Rated Rotor Speed Rated Slip 1.0 pu 1512 rpm -0.008 Number of Pole Pairs 2 14.74 kN m 1.0 pu (соntinued) Rated Mechanical Tarque

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

See similar textbooks

Similar questions

solve D and E.
The series field winding of a 170 kW, 250 V, 1450 rpm compound generator has a resistance of 0.00306 ohm. What will be the required resistance (in Ohm) of a noninductive diverter that will bypass 27 percent of the total armature current? "
Q8) A two-pole de generator has an armature containing a total of 40 conductors connected in two parallel paths. The flux per pole is 0.02 web, and the speed of the prime mover is 30 rpm. The resistance of each conductor is 0.01 ohm, and the current-carrying capacity of each conductor is 10 A. Calculate: (a) The average generated voltage per path and the generated armature voltage (b) The armature current delivered to an external load (c) The armature resistance (d) The terminal voltage of the generator.
A shunt DC generator has no load terminal voltage of 140V. Armature resistance is 1ohm and shunt field resistance is 50ohm. Calculate the induced voltage.
A three-phase, Y-connected, 220-V synchronous generator has an armature resistance of 0.22 Q/phase and a synchronous reactance of 5.2 Q/phase. At the rated voltage, the generator delivers 9.7-kVA to a load at a power factor of 0.83 lagging. A. Determine the excitation voltage, EA. B. Determine the efficiency of the generator if the friction and windage losses are 0.29 KW and the core losses are 0.45 KW. Ignore stray losses. Insert the efficiency below (numbers only):
With the aid of diagrams describe the basic topology of a wound field 3 phase synchronous generator and outline the relationship between the excitation voltage and rotor current. Q2 (a) (b) Outline with the aid of a suitable diagram the necessary power conversion stages required to interface a small permanent magnet (PM) generator to a stand alone DC system. A star connected 3 phase wound field synchronous generator with a synchronous reactance of 20 N is connected to a 6.6 kV (phase voltage) grid and supplies 1.5 MW at 0.95 lagging power factor at its terminals. Calculate the phase current and resultant voltage across the synchronous reactance (Vxs), and from a scaled phasor diagram graphically determine the required excitation voltage (Eph) and load angle (8). (c) The load is now increased to 3 MW with the Excitation Voltage kept constant. Using the phasor diagram drawn in part (c) graphically determine the resultant phase current, power factor and load angle. (d)
Q2. a) A 415V, 50Hz, 3-phase induction motor supplies its rated power to a load. The rated speed of the motor is 980 rpm. b) Two synchronous generators, G2 and G2, are connected parallelly supplying a load. Generator G1 has a no-load frequency of 50.5 Hz and a slope of 300 MW/Hz. Generator G2 has a no-load frequency of 50.2 Hz and a slope of 500 MW/Hz. The load consumes 250 MW real power. (i) At what frequency does this system operate, and how much power is supplied by each of the two generators? (ii) An additional 100 MW load is added to this power system. What is the new system frequency, and how much power do G1 and G2 supply? (iii) The governor set point of G2 is changed to control system frequency back to 50 Hz. Determine the G2 governor set point.
Show that an alternator running in parallel on constant-voltage and frequency bus-bars has a natural time period of oscillation. Deduce a formula for the time of one complete oscillation and calculate its value for a 5000-kVA, 3-phase, 10,000-V machine running at 1500 r.p.m. on constant 50-Hz bus-bars. The moment of inertia of the whole moving system is 14,112 kg-m2 and the steady short-circuit is five times the normal full-load value. ANSWER IS [1.33 second]
A 2300V, 1000 kVA, 0.8 lagging power factor, two-pole, 60HZ, Wye-connected, synchronous generator has a synchronous reactance of 1.1 ohms and an armature resistance of 0.15 ohms. Its losses due to friction and friction with the air are 24 kW and its core losses are 18 kW. The field circuit has a dc voltage of 200v and a maximum of 10A. The field circuit current is adjustable in the range of 20 to 200 ohms.The magnetization curve of this generator is shown in the figure below.a) How much field current is required for Vt to equal 2300V when the generator is running without load?b) What is the voltage generated intermittently by this machine at nominal conditions?c) How much field current is needed to make the voltage at the terminals Vt=2300V when the generator works in nominal conditions?d) How much power and torque should the prime mover of the generator be able to supply?
State two reasons why a self-excited shunt dc generator may fail to build up a voltage. In each (both) case, briefly explain what can be done to correct the problem. Reason #1 Solution #1 Reason #2 Solution #2
Choose True (V) or false( x) for the following : 4) Reluctance is a measure of "opposition" the magnetic circuit offers to the flux. 5) The rated output power (transformer rating) can be used to determine the turns ratio of the transformer. 6) With the increase in speed of a series DC motor, Both back emf as well as line current increase. 7) When an induction motor is stationary, the stator and rotor windings form the equivalent of a transformer. 8) Increasing the load current for DC Self-excited generator Shunt-Wound Generator will increase the terminal voltage. 9) In lap windings there are as many paths in parallel and used to produce high current, low voltage output. 10) In 3-pahse induction motor, if the rotor speed increases, the rate at which the rotating magnetic field cuts the rotor bars is increasing.
Q/ dc shunt generator 4 pole with 500 condcutor , wave running at speed 1000 r.p.m the resistance of armature and field are 0.02 , 55 ohm and delivering load 43.12kwatt at terminal voltage 220 volt, the iron losses is 750 watt 1-Torque developed by the prime mover .........and The torque developed by the armature ........... when running at 1000 r.p.m * 2-The load current ,,,,,,,,,,,,when the speed drops to 800 r.p.m. if If is unchanged ? * 3-Output of the prime motor…….kwatt * 4-Mechanical , electrical and overall efficiency …….. * 5-Flux per pole is......... wb *